如何证明三角形重心特点

2024-11-24 07:29:02
推荐回答(3个)
回答1:

1)重心是三角形三条中线的交点

2)重心到三角形顶点的距离等于它到对边中点距离的二倍。
3)若三角形三个顶点坐标为(x1,y1)
(x2,y2)
(x3,y3),
则重心坐标为[(x1+x2+x3)/3,(y1+y2+y3)/3]

回答2:

证明:
连结ao并延长,交bc于e,连结de
因为cd是ab边上的中线,点o是三角形abc的重心
所以ae是bc边上的中线
所以ad=db,ce=eb
所以de是三角形abc的中位线
所以ed‖ac,ed=1/2ac,即ed/ac=1/2
所以△oed∽△oac
所以od/oc=ed/ac=1/2
即oc=2od

回答3:

角形
重心,证明,三角形
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